Phys 250 Quantum Optics, Homework #3, Solutions

1 Problem 2.7

We can express any state as a sum of number states:

|ψ〉 =∑n

Cn |n〉

So...

|ψ′〉 = a |ψ〉

= N∑n

Cn√n |n− 1〉

⇒ 〈ψ′|ψ′〉 = N2∑n

C2nn (due to orthogonality of |n〉)

= N2n

⇒ N =1√n

What about n′?

n′ = 〈ψ′| n |ψ′〉

=1

n

∑n

C2nn(n− 1)

=1

n

(⟨n2⟩− n

)=〈n2〉n− 1

This is not equal to n − 1 unless 〈n2〉 = 〈n〉2, or in other words |ψ〉 is a number state to beginwith! This is because a photon has a higher probability of begin absorbed from a higher excitedstate (stimulated absorption).

2 Problem 2.8

It should be obvious that n = 5.

1

Furthermore:

n′ =〈n2〉n− 1

=12(02) + 1

2(102)

5− 1

= 9?

This may seem strange, but the output from the (normalized) annihilation operator is weighted –the ground state ends up with a weight of 0, hence it does not contribute! (Na |ψ〉 → |9〉, so thefinal state only has a term due to the |10〉 part of the wavefunction.)

Extended explanation: Having a process where you “know” a photon is absorbed implies somesort of “post-selection” is going on – this is actually relatively common in experimental quantumoptics. If a photon was definitely absorbed, the state couldn’t have been in the ground state tobegin with; some how this is equivalent to measuring the state before the absorption process andfinding it to be in the excited state, |10〉 (and ignoring the cases where it wasn’t).

So... how does this work in theory? To conserve energy, an annihilation operator for one modeshould be coupled to a creation operator for another state of the same energy (or several creationoperators for different states whose energy adds up to the original state). A physically sensibleoperator should like more like this:

O ∼ 1 +N(a†b+ a†b

),

where b is the annihilation operator for the second mode, which was previously ignored. The a†bterm is needed for time-reversal symmetry, although it doesn’t really matter here. Our initial statefor the complete system must be something like this:

|ψ〉 =1√2

(|0〉a + |10〉a)⊗ |0〉b ,

where we assume the second mode is in the ground state. We then apply the operator:

O |ψ〉 ∼ (|0〉a + |10〉a)⊗ |0〉b +N√

9 |9〉a ⊗ |1〉b

If we wish to consider only states where we know a photon was absorbed from mode A, we canmeasure mode B and post-select for the states in which it is excited, which looks like:

〈1|b O |ψ〉 ∼ |9〉a

Thus, we recover the effect of an annihilation operator by using a physical operator plus post-selection on a measurement. This implies, by necessity, that we are actually ignoring the part ofthe wavefunction where nothing happened.

2

3 Problem 2.10

∑n

n(n− 1) . . . (n− r + 1)Pn =∞∑n=r

n!

(n− r)!Pn

=1

1 + n

∞∑n=r

n!

(n− r)!xn; from 2.145 with x =

n

1 + n

=xr

1 + n

∞∑m=0

(m+ r)!

m!xm; where m = n− r

=xr

1 + nr! (1− x)−(r+1)

=

(r!

1 + n

)(n

1 + n

)r(1 + n)r+1

= r!nr

4 Problem 2.12

For the mixed state, start with (2.229):

P(φ) =1

2π〈φ| ρ |φ〉

=1

4π

(〈φ|0〉 〈0|φ〉︸ ︷︷ ︸

1

+ 〈φ|1〉 〈1|φ〉︸ ︷︷ ︸1

)=

1

2π

For the superposition, start with (2.226):

P(φ) =1

2π

∣∣∣∣∣∑n

e−inφCn

∣∣∣∣∣2

Cn =1√2

{1, eiφ, 0 . . .

}⇒ P(φ) =

1

4π

∣∣1 + ei(θ−φ)∣∣2

=1 + cos(θ − φ)

2π

The mixed state has a flat distribution – this is expected because it is just a statistical mixtureof two number states, and number states have a flat phase distribution (no phase information).A superposition of two number states, however, does have phase information, which results in alumped distribution.

3

5 Problem 2.13

The density operator of the thermal state in the number basis is given by (2.138):

ρTh =∑n

Pn |n〉 〈n| ,

but this is just a statistical mixture of number states. As with the last problem, this means thatthe distribution is flat: P(φ) = 1

2π. Of course it has to be, there should be no reason for a thermal

state to prefer a specific phase!

6 Problem 3.1

a† |α〉 = α |α〉

=∑n

Cn√n+ 1 |n+ 1〉

= α∑n

Cn |n〉

⇒ Cn√n+ 1 = αCn+1

⇒ Cn+1 =

√n+ 1

αCn

Which means that Cn monotonically increases with n. This means the state is not normalizableand hence non-physical.

7 Problem 3.4

The easiest way to prove this is to calculate the partial derivative.

4

∂

∂α|α〉 =

∂

∂αe−

αα∗2

∑n

αn√n!|n〉

= −α∗

2|α〉+ e−

αα∗2

∑n

nαn−1

√n!|n〉 (Mathematical reminder:

∂

∂αα∗ = 0)

= −α∗

2|α〉+ e−

αα∗2

∑n

(n+ 1)αn√(n+ 1)!

|n+ 1〉 (Because the n=0 term is 0!)

= −α∗

2|α〉+ e−

αα∗2

∑n

√n+ 1αn√n!

|n+ 1〉

= −α∗

2|α〉+ a† |α〉

∂

∂α〈α| = −α

∗

2〈α|

⇒ ∂

∂α|α〉 〈α| =

(a† − α∗

)|α〉 〈α|

⇒ a† |α〉 〈α| =(α∗ +

∂

∂α

)|α〉 〈α|

The proof of second identity is nearly identical.

8 Problem 3.5

First we calculate⟨X⟩

and⟨X2⟩

:

X1 =1

2

(a+ a†

)⟨X1

⟩α

=1

2〈α| a+ a† |α〉

=1

2(α + α∗) 〈α|α〉

= Re (α)

5

⟨X2

1

⟩α

=1

4〈α| a2 + aa† + a†a+ a†2 |α〉

=1

4〈α| a2 + 2a†a+ 1 + a†2 |α〉

=1

4

(α2 + 2αα∗ + α∗2 + 1

)=

1

4

[(α + α∗)2 + 1

]=

1

4+ Re (α)2

⇒⟨(

∆X1

)2⟩α

=⟨X2

1

⟩α−⟨X1

⟩2

α=

1

4

X2 =1

2i

(a− a†

). . .⟨

X2

⟩α

= Im (α)

⟨X2

2

⟩α

= −1

4〈α| a2 − aa† − a†a+ a†2 |α〉

= −1

4〈α| a2 − 2a†a− 1 + a†2 |α〉

. . .

=1

4+ Im (α)2

⇒⟨(

∆X2

)2⟩α

=⟨X2

2

⟩α−⟨X2

⟩2

α=

1

4

6